The Wye (also know as Star - especially in the motor rewind industry) is a 4-wire system which provides two different supply voltages. The center-point of the Wye is the system neutral and is usually solidly grounded. Where it is desirable to limit the phase-to-ground fault magnitude the center-point of the Wye may be connected to ground through and neutral grounding resistor or a current limiting reactor. Because the system is tied to ground it is easy to provide system ground fault protection. Three-phase loads can be connected phase-to-phase and singlephase loads can be connected from any phase to the system neutral. On a wye system, the phase unbalance currentis carried by the system neutral. On a Wye system the line current is equal to the phase current i.e. ( ILine 1 = IPhase A) and the line-to-line voltage is equal to the vector sum of two individual phase voltages i.e. (E Line1-2 = E PhaseA + E PhaseB' ). In a Wye system the phase-to-phase voltage is 1.732 x the phase-to-ground voltage. Some typical Wye system voltages are: 120/208Y, 277/480Y, 2400/4160Y, 4160/7200Y, 7200/12470Y, 7620/13200Y,and 19920/34500Y
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Just consider the a DC voltage as the function y = (DC value), lets call the DC value 1 volt. Then the integral of y=1 is y=x. The op amp will create a ramp that is representative of y=x until it saturates. Depending on the op amp, it will most likely saturate somewhere around Vcc - 2V. As soon as it hits this rail it will level off.
Vpp is Peak-to-Peak voltage, in other words, in AC voltage, the peak-to-peak voltage is the potential difference between the lowest trough in the AC signal to the highest. Assuming the reference to the voltage is zero, Vpp would be twice the peak voltage (between zero and either the highest or lowest point in the AC waveform). Vrms is the Root Mean Square voltage, think of it as sort of an average (it's not quite that simple). For a sine wave, the RMS voltage can be calculated by y=a*sin(2ft) where f is the frequency of the signal, t is time, and a is the amplitude or peak value.
It's a description of the some of the electrical properties of the transformer. This is a Delta connected highside (the D), wye connected lowside (Y) that is grounded (N) and there is a 330 degree phase shift between the highside and the lowside (the low voltage is leading the high voltage by 30 degrees).
Voltage is the potential difference between the source & any point in the circuit. The forward voltage is the voltage drop across the diode if the voltage at the anode is more positive than the voltage at the cathode (if you connect + to the anode). Voltage drop means, amount of voltage by which voltage across load resistor is less then the source voltage.
by using Y-Y(Star) transformer..... bcoz in star VL = root 3* Vph
A: A DELTA transformer is a 1:1 voltage transfer delta to Y IS 1:2 voltage transfer. That is for 3 phase system, If the phases are not exactly matched or the voltage is not exactly right then on a Y setup there will be circulating current at the common node.
connect the neutral point of the star (secondary) to u phase of primary, and now apply voltage to primary and measure the voltage between V&R, W&B, V&Y and V&B. when you measure voltage b/w 1. V&R the voltage must be maximum, 2. W&B Should be minimum and 3. voltage between V&Y and V&B should be same U.V&W are primary (Delta) R,Y&B are secondary (Star)
30 MV open circuit
You should look at the transformer vector diagram, or the type of transformer will tell you (something like YnD1, for example). If it's a Y/Y type transformer, the phase angle will be zero. If it is a D/D, it will be zero. If it's a Y/D, or D/Y, it could be +30, -30 (usually). I have seen a few strange cases where a Y/D was +150 degrees.
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You generally plot a graph of Voltage v/s Current (with voltage on Y axis and current on the X axis). The graph will not be a straight line.
Just consider the a DC voltage as the function y = (DC value), lets call the DC value 1 volt. Then the integral of y=1 is y=x. The op amp will create a ramp that is representative of y=x until it saturates. Depending on the op amp, it will most likely saturate somewhere around Vcc - 2V. As soon as it hits this rail it will level off.
In vcd, input voltage controls the output current.e.g.jfet In ccd , input current controls the output current. e.g. bjt
I assume you are asked to find peak voltage on a graph. If so its simply the number of divisions times four volts for the highest point on the graph.
f your supply is a Y connected transformer (4 wires COM) with a phase to phase voltage of 380 volts (voltage between any two of the hot wires) , utilizing a true Y connection to your load (connections of one hot lead and neutral for each phase to your load will give a voltage of 380 รท sqrt 3 = 380 รท 1.732 = 220 volts.
An ideal zener diode will have zero reverse current while the reverse voltage is less than the zener voltage. Once the voltage rises above the zener voltage, the maximum reverse current will become infinite (the device will become a short). On a graph with voltage along the X axis and current along the Y axis, this would be represented by a straight vertical line crossing through the zener voltage. A practical zener diode has a monotonic change from zero current at zero volts, rising gradually as the voltage approaches the zener voltage from below, then rising sharply as the voltage is around the zener voltage. This means that with reverse voltage applied even slightly below the zener voltage there will be some current flow. This can be a problem in some circuits if not understood and accounted for.